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3 years ago

1. What is the most precise name for quadrilateral ABCD with vertices A(-2,4), B(5,6), C(12,4), and D(5,2)?

Mathematics

2 answers:

zaharov [31]3 years ago

7 0

The first one is B because of the first attachment. The second one is A because of the second attachment. The third one is C because of the expression used to solve for sum of angles:
(n-2)180
(8-2)180
1080
Hope this helps!

lys-0071 [83]3 years ago

5 0

Answer:

1) b

2)a

3)b

Step-by-step explanation:

1)Consider AC and BD. Both have common mid point as (5,4) and (5,4)

Hence diagonals bisect each other. Slope of AC=0 and slope of BD = -4/0

i.e. one is horizontal and other is vertical. Hence diagonals bisect at right angles. So rhombus.

2) ABCD has adjacent sides equal and diagonals bisect each other at right angles.

Slope of AC =1 and BD =-1 that is product =-1. Kites are perpendicular.

So kite

3) Any polygon irrespective of number of sides, the exterior angles will add upto 360 degrees.

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-3/10 + 4/15<br> Solve the above problem

Firlakuza [10]

Answer:

-1/30

Step-by-step explanation:

find the lcm of 10 and 15

then solve it

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2 years ago

How many plants spaced every 6 in. are needed to surround a circular walkway within a 25 ft. radius?

e-lub [12.9K]

Consider the following formula to find the circumference of circle
C= 2^_r
here r is radius of circle
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3 years ago

Verify that the points are the vertices of a parallelogram, and find its area. A(1, 1, 3), B(2, −5, 6), C(4, −2, −1), D(3, 4, −4

almond37 [142]

Answer:

Area = 71.3 sq unit

Step-by-step explanation:

a) We will use properties of parallelogram to verify the given vertices.

Property: Have a pair of parallel opposite sides

Hence, vectors AB AD BD BC CD AC. A pair should be parallel:

vectors

AB = A - B = (1, 1 , 3) - ( 2 , -5 , 6 ) = (-1 , 6 , -3)

AD = A - D = (1, 1 , 3) - (3, 4 , -4) = (-2 , -3 , 7)

BC = B - C = (2 , -5 , 6) - (4, -2 , -1) = (-2 ,-3 , 7)

CD = C - D = (4, -2 , -1) - (3, 4 , -4) = (1 , -6 , 3)

Hence we can see that AB //CD and AD // BC as their unit vector co-efficents are identical/scalar multiple of each other.

Equation of line AB = (1,1,3) + t*(-1 , 6 , -3 )

Point E = (1-t , 1+6t , 3 -3t) ... denotes an position of point E on line AB.

Choose a point on line CD, lets suppose C = ( 4 , -2 , -1 )

Vector EC = ( 1-t , 1+6t , 3-3t ) - (4 , -2 , -1 ) = (-3 -t , 3 +6t , 4 -3t)

For EC to be perpendicular to AB then dot product of AB . EC = 0

Hence,

EC . AB = -1 * (-3-t) -2*(3+6t) -3*(4-3t) = 0

3 + t -6 -12t -12+9t = 0

-2t-3 = 0

t = -3/2

Vector EC = (-3 -t , 3 +6t , 4 -3t) = (-1.5 , -6 , 8.5)

Area = magnitude (AB) * magnitude (EC)

Area = sqrt ((-1)^2 + 6^2 + (-3)^2) * sqrt ((-1.5)^2 + 6^2 + (8.5)^2)

Area = sqrt (46) * sqrt (442) / 2

Area = 71.3 unit^2

6 0

2 years ago

Jaylen earns $4250 each month at his job. At this rate, how much will he

dezoksy [38]

Answer:

Step-by-step explanation:

<h2>18 months in 1 $\frac{1}{2}$ years.</h2><h2>4250 x 18 = 76,500</h2>

4 0

1 year ago

Solve the following system of equations:

Vladimir79 [104]

Answer:

Step-by-step explanation:

We can solve this by either graphing and finding ther point the lines intersect, or mathematically, I'll do both.

<u>Graphing:</u>

<u>Mathematically:</u>

−2x + 4y = 6

y = 2x + 3

See the attached graph. The lines intersect at (-1,1)

--------------------

I'll rearrange the first equation (to make it easier for me):

−2x + 4y = 6

4y = 2x + 6

y = (1/2)x + 1.5

Now lets substitute the second equation into the first so that we can eliminate y:

y = 2x + 3

[(1/2)x + 1.5] = 2x + 3

- (3/2)x = (3/2)

x = -1

If x = -1:

y = 2(-1) + 3

y = 1

The solution is x = -1 and y = 1, or (-1,1)

=================

Both approaches give us (-1,1), the solution to the system of equations. It is the only point that satisfies both equations.

3 0

1 year ago